This is my first question on the site and I'm sure there'll be many more. But for now to the point.

I am really having trouble understanding the proper use of roots and exponents to try to solve inequalities. The thing is that I have noticed that using such operations on many occasions do not produce an equivalent inequation.

My problems started with the following excercise $\sqrt{8x^{2}+22x+15} > 4x+3$

I first considered that $\ 8x^{2}+22x+15\geq 0$, so I know that root expression will be a real number as long as $\ x \leq-3/2$ OR $\ x \geq-5/4$

My attempted solutions:

1) Square both sides to obtain a second-degree polynomial on each side, simplify and solve.

$$\ \big( \sqrt{8x^{2}+22x+15} \big) ^{2} > \big(4x+3\big) ^{2} $$

RESULT: obtained the wrong solution and I'm under the impression that squaring doesn't produce an equivalent inequality.

2) Subtract $\ 4x + 3 $ on both sides and try to eliminate the square roots by squaring both sides.

If I attempt this $\ 24x^2+46x+12 > 2\big(4x+3\big)\sqrt{8x^{2}+22x+15} $ and square both sides to eliminate the square root I will get the wrong answer in the end since I know that this does not produce an equivalent inequation. (Tried solving like this before so I kinda figured it was a wrong move.)

I would like an insight as to how to solve these kind of problems and what is the correct way to do them. What mathematical facts or rules am I missing here? How should I treat roots and exponents in general?

3 Answers
3

The underlying question here, I beleive, is what's called monotony, which basically means increasing or decreasing. Consider a function a real-valued function $f$, that takes real variables, with the property that: $$\text{For real numbers $x,y$ of a subset A} \\ x<y \Rightarrow f(x) < f(y)$$
Then we say that $f$ is increasing (in $\text{A}$) This happens, for example, with $f\mid f(x)=2x$. On the other hand, take $g \mid g(x)=x^2$, and the numbers $-3,2$. Obviously $-3<-2$, but $g(-3) \not< g(2)$. Actually if $x<0<y$, then $g(x)\not <g(y)$. This is why $$\sqrt{8x^{2}+22x+15} > 4x+3\not\Rightarrow \ \big( \sqrt{8x^{2}+22x+15} \big) ^{2} > \big(4x+3\big) ^{2},
$$ because $4x+3$ can be negative for, say $x=-2$, which is in the domain you have found that the square root is real (and therefore the expression makes sense). In fact, $4x+3$ is negative if and only if $x<\large\frac{-3}{4}$. Also notice that $x>\large\frac{-3}{4}\Rightarrow $ $x \ge \large\frac{-5}{4}$, and so, for all $x>\large\frac{-3}{4}$ $$\big( \sqrt{8x^{2}+22x+15} \big) ^{2} > \big(4x+3\big) ^{2}
$$
You can continue as before, taking into account that the result applies for the mentioned $x$'s. Say you obtain $x\in P\subset\mathbb R$ ($P$ some subset of real numbers).

For $x < \frac{-3}{4}$ (for $x < \frac{-3}{2}$ to be precise, because the expression is only defined here), the expression holds always. There the original inequality implies one of two situations: $$\left\{\begin{align} x &<\frac{-3}{2} \\ &\mathrm {or} \\ x&\in P\subset\mathbb R\end{align}\right.$$
This should be the final result, correct me if I'm wrong!

As for roots and exponents in general, or any function we "apply" to an inequality (I use quotations because really we are just taking into account the definition of increasing or decreasing, rather than "applying" the function), we just have to find the subsets where the function increases or decreases (decreasing has the same definition, replacing the second "$<$" with "$>$"). Then the procedure will yield, for each subset, a different subset of real numbers in which the inequality holds (could be empty sets). For example, for natural $n$, $h\mid h(x)=x^n$ is increasing in all of $\mathbb R$ if and only if $n$ is odd, and behaves as $x^2$ for even $n$. Try figuring out similar properties for positive rational exponents and positive $x$ (hint: write $x^{p/q}$ as $(x^p)^{\frac1q}$ and use the rules for natural exponents shown before, and think whether the $q$-th root of a number is increasing or decreasing (the function that assigns the $q$-th root). Hope I helped out!

Well, you will have to square the inequation (at least) once, so prefer to choose attempt 1).

It is indeed true that squaring an equation is not an equivalent step, however, now it is of the form $A>B$ with $A\ge 0$ (as $A=\sqrt{\text{something}}$). So, either $B<0$ -- in which case the inequality holds --, or both $A,B$ are nonnegative, and among nonnegative numbers squaring is an equivalent step: